例子:

absolute="/foo/bar"
current="/foo/baz/foo"

# Magic

relative="../../bar"

我如何创造魔法(希望不是太复杂的代码…)?


当前回答

我把你的问题作为一个挑战,用“可移植的”shell代码来编写它,即。

考虑到POSIX外壳 没有数组之类的bashisms 避免像打瘟疫一样打外部电话。脚本中没有一个分叉!这使得它非常快,特别是在有显著分叉开销的系统上,比如cygwin。 必须处理路径名中的glob字符(*,?,[,])

它运行在任何POSIX兼容shell (zsh, bash, ksh, ash, busybox,…)上。它甚至包含一个测试套件来验证其操作。路径名的规范化留作练习。: -)

#!/bin/sh

# Find common parent directory path for a pair of paths.
# Call with two pathnames as args, e.g.
# commondirpart foo/bar foo/baz/bat -> result="foo/"
# The result is either empty or ends with "/".
commondirpart () {
   result=""
   while test ${#1} -gt 0 -a ${#2} -gt 0; do
      if test "${1%${1#?}}" != "${2%${2#?}}"; then   # First characters the same?
         break                                       # No, we're done comparing.
      fi
      result="$result${1%${1#?}}"                    # Yes, append to result.
      set -- "${1#?}" "${2#?}"                       # Chop first char off both strings.
   done
   case "$result" in
   (""|*/) ;;
   (*)     result="${result%/*}/";;
   esac
}

# Turn foo/bar/baz into ../../..
#
dir2dotdot () {
   OLDIFS="$IFS" IFS="/" result=""
   for dir in $1; do
      result="$result../"
   done
   result="${result%/}"
   IFS="$OLDIFS"
}

# Call with FROM TO args.
relativepath () {
   case "$1" in
   (*//*|*/./*|*/../*|*?/|*/.|*/..)
      printf '%s\n' "'$1' not canonical"; exit 1;;
   (/*)
      from="${1#?}";;
   (*)
      printf '%s\n' "'$1' not absolute"; exit 1;;
   esac
   case "$2" in
   (*//*|*/./*|*/../*|*?/|*/.|*/..)
      printf '%s\n' "'$2' not canonical"; exit 1;;
   (/*)
      to="${2#?}";;
   (*)
      printf '%s\n' "'$2' not absolute"; exit 1;;
   esac

   case "$to" in
   ("$from")   # Identical directories.
      result=".";;
   ("$from"/*) # From /x to /x/foo/bar -> foo/bar
      result="${to##$from/}";;
   ("")        # From /foo/bar to / -> ../..
      dir2dotdot "$from";;
   (*)
      case "$from" in
      ("$to"/*)       # From /x/foo/bar to /x -> ../..
         dir2dotdot "${from##$to/}";;
      (*)             # Everything else.
         commondirpart "$from" "$to"
         common="$result"
         dir2dotdot "${from#$common}"
         result="$result/${to#$common}"
      esac
      ;;
   esac
}

set -f # noglob

set -x
cat <<EOF |
/ / .
/- /- .
/? /? .
/?? /?? .
/??? /??? .
/?* /?* .
/* /* .
/* /** ../**
/* /*** ../***
/*.* /*.** ../*.**
/*.??? /*.?? ../*.??
/[] /[] .
/[a-z]* /[0-9]* ../[0-9]*
/foo /foo .
/foo / ..
/foo/bar / ../..
/foo/bar /foo ..
/foo/bar /foo/baz ../baz
/foo/bar /bar/foo  ../../bar/foo
/foo/bar/baz /gnarf/blurfl/blubb ../../../gnarf/blurfl/blubb
/foo/bar/baz /gnarf ../../../gnarf
/foo/bar/baz /foo/baz ../../baz
/foo. /bar. ../bar.
EOF
while read FROM TO VIA; do
   relativepath "$FROM" "$TO"
   printf '%s\n' "FROM: $FROM" "TO:   $TO" "VIA:  $result"
   if test "$result" != "$VIA"; then
      printf '%s\n' "OOOPS! Expected '$VIA' but got '$result'"
   fi
done

# vi: set tabstop=3 shiftwidth=3 expandtab fileformat=unix :

其他回答

kasku和Pini的答案略有改进,空格更好,允许传递相对路径:

#!/bin/bash
# both $1 and $2 are paths
# returns $2 relative to $1
absolute=`readlink -f "$2"`
current=`readlink -f "$1"`
# Perl is magic
# Quoting horror.... spaces cause problems, that's why we need the extra " in here:
relative=$(perl -MFile::Spec -e "print File::Spec->abs2rel(q($absolute),q($current))")

echo $relative

test.sh:

#!/bin/bash                                                                 

cd /home/ubuntu
touch blah
TEST=/home/ubuntu/.//blah
echo TEST=$TEST
TMP=$(readlink -e "$TEST")
echo TMP=$TMP
REL=${TMP#$(pwd)/}
echo REL=$REL

测试:

$ ./test.sh 
TEST=/home/ubuntu/.//blah
TMP=/home/ubuntu/blah
REL=blah

这里的答案并不是每天都能用的。由于在纯bash中很难正确地做到这一点,我建议以下可靠的解决方案(类似于注释中的一个建议):

function relpath() { 
  python -c "import os,sys;print(os.path.relpath(*(sys.argv[1:])))" "$@";
}

然后,你可以得到基于当前目录的相对路径:

echo $(relpath somepath)

或者你可以指定路径相对于给定的目录:

echo $(relpath somepath /etc)  # relative to /etc

一个缺点是这需要python,但是:

它在任何python >= 2.6中工作相同 它不要求文件或目录存在。 文件名可以包含更广泛的特殊字符。 例如,如果文件名包含 空格或其他特殊字符。 它是一个单行函数,不会使脚本混乱。

注意,包含basename或dirname的解决方案不一定更好,因为它们要求安装coreutils。如果有人有可靠而简单的纯bash解决方案(而不是令人费解的好奇心),我会感到惊讶。

在bash中:

realDir=''
cd $(dirname $0) || exit
realDir=$(pwd)
cd -
echo $realDir
$ python -c "import os.path; print os.path.relpath('/foo/bar', '/foo/baz/foo')"

给:

../../bar